Prove $\sum_{r=2}^n {n \choose r} r(r-1) = n(n-1)2^{n-2}$ for $n\geq 2$ How to prove following:
$$\sum_{r=2}^n {n \choose r} r(r-1) = n(n-1)2^{n-2}$$ for $n\geq 2$
Thanks!!
 A: Question: On how many ways can we choose a group in a set of $n$ people and then president and then vicepresident?
Well we can first a group of $r$ people, that is ${n\choose r}$, for every $r\leq n$, and then president among them, so we have $r$ choises and then $r-1$ choises for V.P. Suming (we can sum from $0$) for all $r$ we get:
$$\sum_{r=0}^{n} r(r-1)C_{n}^{r} $$
On the other hand we can first choose a president among all people, so we have $n$ posibilities and then V.P. for who we have $n-1$ choises and then we choose any set in set of $n-2$ people, for that we have $2^{n-2}$ choises, so:
$$n(n-1)2^{n-2}$$ and this is the answer to your question.
A: Hint: Take $f(x)=(1+x)^n$ and consider $f''(1)$.
A: Hint:
For $r(r-1)\ne0$
$$r(r-1)\binom nr=r(r-1)n(n-1)\dfrac{(n-2)!}{r(r-1)\cdot(r-2)!\cdot\{n-2-(r-2)\}!}=n(n-1)\binom{n-2}{r-2}$$
Now in 
$$(a+b)^m=\sum_{r=0}^m\binom mr a^{m-r}b^r$$
put $a=b=1, m=n-2$
Some observations :


*

*$$\sum_{r=2}^n\binom nrr(r-1)=\sum_{r=0}^n\binom nrr(r-1)$$

*The proposition trivially holds true for $n=0,1$
